Problem: $f(x)=3x^4-16x^3+24x^2+48$. On which intervals is the graph of $f$ concave down? Choose 1 answer: Choose 1 answer: (Choice A, Incorrect) Incorrect $x<-2$ and $x>-\dfrac{2}{3}$ (Choice B, Incorrect) Incorrect $x<\dfrac{2}{3}$ and $x>2$ (Choice C, Incorrect) Incorrect $x>0$ only (Choice D, Checked, Correct) Correct (selected) $\dfrac{2}{3}<x<2$ only
Answer: We can analyze the intervals where $f$ is concave up/down by looking for the intervals where its second derivative $f''$ is positive/negative. This analysis is very similar to finding increasing/decreasing intervals, only instead of analyzing $f'$, we are analyzing $f''$. The second derivative of $f$ is $f''(x)=12(3x-2)(x-2)$. $f''(x)=0$ for $x=\dfrac{2}{3},2$. Since $f''$ is a polynomial, it's defined for all real numbers. Therefore, our points of interest are $x=\dfrac{2}{3}$ and $x=2$. Our points of interest divide the number line into three intervals: $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $x<\frac{2}{3}$ $\frac{2}{3}<x<2$ $x>2$ Let's evaluate $f''$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f''(x)$ Verdict $x<\dfrac{2}{3}$ $x=0$ $f''(0)=48>0$ $f$ is concave up $\cup$ $\dfrac{2}{3}<x<2$ $x=1$ $f''(1)=-12<0$ $f$ is concave down $\cap$ $x>2$ $x=3$ $f''(3)=84>0$ $f$ is concave up $\cup$ In conclusion, the graph of $f$ is concave down over the interval $\dfrac{2}{3}<x<2$ only.